Question

In the circuit shown in Fig. 1, V = 120 V, R1 = 402, R2 = 209, and R3 = R1 = 109 a) What is the equivalent resistance of this circuit? b) What is the current through 13? c) What is the potential difference across R2? d) What is the total power used by this circuit? w RE ww R, RA Figure 1: Problem 1

please show calculations and explanation when necessary

Answer 1

Solution :

Here we have :

V = 120 V

R₁ = 40 Ω

R₂ = 20 Ω

R₃ = 10 Ω

R₄ = 10 Ω

.

Here, R₃ and R₄ are connected in series combination:

∴ R₃₄ = R₃ + R₄ = 10 Ω + 10 Ω = 20 Ω

.

And, the combination of R₃ and R₄ is connected in parallel combination with R₂.

Thus : R₂₃₄ = R₂ R₃₄/ (R₂ + R₃₄) = (20 Ω)(20 Ω) / (20 Ω + 20 Ω) = 10 Ω

.

Therefore, Equivalent resistance of the circuit will be : R_eq = R₁ + R₂₃₄ = 40 Ω + 10 Ω = 50 Ω

∴ R_eq = 50 Ω

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So, Total current flowing through the circuit will be : I = V / R_eq = (120 V) / (50 Ω) = 2.4 A

Thus : I₁ = I = 2.4 A

And, V₁ = I₁ R₁ = (2.4 A)(40 Ω) = 96 V

.

Thus : V₂ = V₃₄ = V - V₁ = 120 V - 96 V = 24 V

∴ V₂ = 24 V

So, I₃ = I₄ = V₃₄ / R₃₄ = (24 V) / (20 Ω) = 1.2 A

∴ I₃ = 1.2 A

.

And, Power used by the circuit : P = V I = (120 V)(2.4 A) = 288 W

∴ P = 288 W

.

Answers :

Part (a) Answer : R_eq = 50 Ω

Part (b) Answer : I₃ = 1.2 A

Part (c) Answer : V₂ = 24 V

Part (d) Answer : P = 288 W